Mathematics for the interested outsider

The Determinant of the Adjoint

It will be useful to know what happens to the determinant of a transformation when we pass to its adjoint. Since the determinant doesn’t depend on any particular choice of basis, we can just pick one arbitrarily and do our computations on matrices. And as we saw yesterday, adjoints are rather simple in terms of matrices: over real inner product spaces we take the transpose of our matrix, while over complex inner product spaces we take the conjugate transpose. So it will be convenient for us to just think in terms of matrices over any field for the moment, and see what happens to the determinant of a matrix when we take its transpose.

Okay, so let’s take a linear transformation and pick a basis to get the matrix

and the formula for the determinant reads

and the determinant of the adjoint is

where we’ve now taken the transpose.

Now for the term corresponding to the permutation we can rearrange the multiplication. Instead of multiplying from to , we multiply from to . All we’re doing is rearranging factors, and our field multiplication is commutative, so this doesn’t change the result at all:

But as ranges over the symmetric group , so does its inverse . So we relabel to find

and we’re back to our formula for the determinant of itself! That is, when we take the transpose of a matrix we don’t change its determinant at all. And since the transpose of a real matrix corresponds to the adjoint of the transformation on a real inner product space, taking the adjoint doesn’t change the determinant of the transformation.

What about over complex inner product spaces, where adjoints correspond to conjugate transposes? Well, all we have to do is take the complex conjugate of each term in our calculation when we take the transpose of our matrix. Then carrying all these through to the end as we juggle indices around we’re left with

The determinant of the adjoint is, in this case, the complex conjugate of the determinant.

I’m not sure offhand what you mean by “by taking alternating products”. You mean the way I defined it originally? It should be possible to prove that in a basis-free manner, but this is the quick and dirty method that I’m going to need before tomorrow’s post.

The problem is that although we can get a canonical basis of , given one of , we don’t necessarily have a canonical bilinear form on (as far as I know, at least). Still, a functorial/basis-free proof would be nice.

[…] of matrices over more general fields). So now that we’ve got information about how the determinant and the adjoint interact, we can see what happens when we restrict the determinant homomorphism to these subgroups […]

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